chứng minh rằng 2+2^2+2^3+…+2^95+2^96 chia hết 21 26/11/2021 Bởi Lydia chứng minh rằng 2+2^2+2^3+…+2^95+2^96 chia hết 21
Đáp án: Đặt tổng : `2+2^2+2^3+…+2^95+2^96 = A` `=> A = 2+2^2+2^3+…+2^95+2^96` `=> A = (2+2^2+2^3+2^4+2^5+2^6)+…..+(2^{91}+2^{92}+2^{93}+2^{94}+2^{95}+2^{96})` `=> A=2×(1+2+2^2+2^3+2^4+2^5)+…+2^{91}×(1+2+2^2+2^3+2^4+2^5)` `=>A=(1+2+2^2+2^3+2^4+2^5)×(2+…+2^{91})` `A=63.(2+…+2^{91}) vdots 21` Vì `63.(2+…+2^{91}) vdots 21 => A vdots 21(đpcm)` Giải thích các bước giải: Bình luận
Tham khảo `A=2+2^2+2^3+2^4+2^5+2^6+…..+2^{91}+2^{92}+2^{93}+2^{94}+2^{95}+2^{96}` `A=(2+2^2+2^3+2^4+2^5+2^6)+…..+(2^{91}+2^{92}+2^{93}+2^{94}+2^{95}+2^{96})` `A=2×(1+2+2^2+2^3+2^4+2^5)+…+2^{91}×(1+2+2^2+2^3+2^4+2^5)` `A=(1+2+2^2+2^3+2^4+2^5)×(2+…+2^{91})` `A=63×(2+…+2^{91})` Vì `63 \vdots 21` `⇒A \vdots 21` Bình luận
Đáp án:
Đặt tổng : `2+2^2+2^3+…+2^95+2^96 = A`
`=> A = 2+2^2+2^3+…+2^95+2^96`
`=> A = (2+2^2+2^3+2^4+2^5+2^6)+…..+(2^{91}+2^{92}+2^{93}+2^{94}+2^{95}+2^{96})`
`=> A=2×(1+2+2^2+2^3+2^4+2^5)+…+2^{91}×(1+2+2^2+2^3+2^4+2^5)`
`=>A=(1+2+2^2+2^3+2^4+2^5)×(2+…+2^{91})`
`A=63.(2+…+2^{91}) vdots 21`
Vì `63.(2+…+2^{91}) vdots 21 => A vdots 21(đpcm)`
Giải thích các bước giải:
Tham khảo
`A=2+2^2+2^3+2^4+2^5+2^6+…..+2^{91}+2^{92}+2^{93}+2^{94}+2^{95}+2^{96}`
`A=(2+2^2+2^3+2^4+2^5+2^6)+…..+(2^{91}+2^{92}+2^{93}+2^{94}+2^{95}+2^{96})`
`A=2×(1+2+2^2+2^3+2^4+2^5)+…+2^{91}×(1+2+2^2+2^3+2^4+2^5)`
`A=(1+2+2^2+2^3+2^4+2^5)×(2+…+2^{91})`
`A=63×(2+…+2^{91})`
Vì `63 \vdots 21`
`⇒A \vdots 21`